Chapter 9: Control of Traction Motors

Example 9.1, Page 268

In [3]:
#Variable declaration
V=600.;# in volts
I=350.;#in A
Ts=20.;# in sec
R=0.15;# in ohm

#Calculations&Results
E_bse=(V/2)-(I*R);
E_bp=V-(I*R);
Tse=(E_bse/E_bp)*Ts;
Tp=Ts-Tse;
Vd=V-(2*I*R);
Ed1=(Vd/2)*I*(Tse/3600);
Ed2=((V/2)/2)*2*I*(Tp/3600);
El=(Ed1+Ed2)*10**-3;
print "part (a)"
print "Energy lost in starting rhestat,El(kWh) = %.4f"%El
El_1=(2*(I**2)*R*Ts)/(3600*1000);
print "part (b)"
print "Energy lost in  motors,El(kWh) = %.3f"%El_1
#answer is wrong in  part b in the textbook
Et=((V*I*Tse)+(2*V*I*Tp))/(3600*1000);
print "part (c)"
print "Total Energy,Et(kWh) = %.3f"%Et

part (a)
Energy lost in starting rhestat,El(kWh) = 0.5372
part (b)
Energy lost in  motors,El(kWh) = 0.204
part (c)
Total Energy,Et(kWh) = 1.806

Example 9.2, Page 269

In [6]:
#Variable declaration
V=600.;# in volts
I=300.;#in A
Ts=15.;# in sec
R=0.1;# in ohm

#Calculations&Results
E_bse=(V/2)-(I*R);
E_bp=V-(I*R);
Tse=(E_bse/E_bp)*Ts;
Tp=Ts-Tse;
Vd=V-(2*I*R);
Ed1=(round((Vd/2)*I*(Tse/3600))*10**-3);#
print "part (i)"
print "rheostatic in series,Ed1(kWh) = %.2f"%Ed1
Ed2=((V/2)/2)*2*I*(Tp/3600)*10**-3;
print "rheostatic in parallel,Ed2(kWh) = %.3f"%Ed2
Vm=29;# in kmph
alfa=Vm/Ts;
S=alfa*Tse;
print "part (ii)"
print "Speed at the end of series period,S(km/h) = %.1f"%S

part (i)
rheostatic in series,Ed1(kWh) = 0.16
rheostatic in parallel,Ed2(kWh) = 0.197
part (ii)
Speed at the end of series period,S(km/h) = 13.7

Example 9.3, Page 270

In [8]:
#Variable declaration
V=600;# in volts
I=200;#in A
Ts=20;# in sec
R=0.1;# in ohm

#Calculations&Results
E_bse=(V/2)-(I*R);
E_bp=V-(I*R);
Tse=(E_bse/E_bp)*Ts;
Tp=Ts-Tse;
Vd=V-(2*I*R);
Mi=((V*I*Tse)/(2*3600))+((V*I*Tp)/3600);
Er=((Vd/4)*I*(Tse/3600))+(((V/2)/2)*I*(Tp/3600));
El=(I**2*R*Ts)/(3600);
Mo=Mi-Er-El;
eta=(Mo/Mi)*100;
print "part (a)"
print "Starting efficiency = %.1f%%"%eta
Vm=80;# in kmph
alfa=Vm/Ts;
S=alfa*Tse;
print "part (b)"
print "speed,S(kmph) = %.2f"%S

part (a)
Starting efficiency = 63.7%
part (b)
speed,S(kmph) = 38.62

Example 9.4, Page 271

In [12]:
#Variable declaration
W=150;# in tonne
We=1.1*W;# in tonnes
Vm=30;#kmph
V=600;# in volts
r=10;# N/tonne
I=300;#in A
R=0.1;# in ohm
Ft=4*15000;# in N
G=1;#gradient in %

#Calculations&Results
alfa=(Ft-(W*r)-(98.1*W*G))/(277.8*We);
Ts=Vm/alfa;
E_bse=(V/2)-(I*R);
E_bp=V-(I*R);
Tse=(E_bse/E_bp)*Ts;
print "part (a)"
print "Duration of starting period,Ts(seconds) = %.1f"%Ts
print "Duration for Series running,Tse(seconds) = %.1f"%Tse
sptr=alfa*Tse;#in kmph
print "part (b)"
print "speed of train at transition in kmph is %.2f"%sptr
sptr=alfa*Tse;#in kmph
rls=((V-(2*I*R))/2)*(2*I)*(Tse/3600);#watts hours
rlp=((V/2)/2)*(4*I)*((Ts-Tse)/3600);#watts hours
tl=rls+rlp;#
print "part (c)"
print "rheostat losses during series operation is %.1f W-hours"%rls
print "rheostat losses during parallel operation is %.f W-hours"%rlp
print "total losses in W-hours is %.1f "%tl

part (a)
Duration of starting period,Ts(seconds) = 31.4
Duration for Series running,Tse(seconds) = 14.9
part (b)
speed of train at transition in kmph is 14.21
part (c)
rheostat losses during series operation is 669.4 W-hours
rheostat losses during parallel operation is 826 W-hours
total losses in W-hours is 1495.9

Example 9.5, Page 272

In [16]:
#Variable declaration
nf=1.; 
n2=1.25*nf; 
of=1; 
of2=nf/n2; 
isef=1; 
ise2=0.66667; 

#Calculations
ia2=(1./ise2); 
idiv=ia2-ise2; 
rdiv=ise2/idiv; 

#Result
print "diverter resistance required as percentage of the field resistance is %.f%%"%(rdiv*100)
#answer is wrong in the textbook

diverter resistance required as percentage of the field resistance is 80%

Example 9.6, Page 272

In [33]:
#Variable declaration
Ia=[60,80,100,120,160,180];# in amperes
sp1=[47.4,40.3,35.8,33.9,29.8,28.5];#in kmph
dpk=[440,700,970,1245,1800,2360];#in kg
sp2=[58.1,50,45,40.3,35,32];#

#Calculations&Results
for i in range(0,6):
    dpk1= ((dpk[i])*(sp1[i]))/(sp2[i]);#
    print "For current = ",Ia[i],"A, speed is ",sp2[i],"kmph and drawbar pull is",round(dpk1),"kg"
    

For current =  60 A, speed is  58.1 kmph and drawbar pull is 359.0 kg
For current =  80 A, speed is  50 kmph and drawbar pull is 564.0 kg
For current =  100 A, speed is  45 kmph and drawbar pull is 772.0 kg
For current =  120 A, speed is  40.3 kmph and drawbar pull is 1047.0 kg
For current =  160 A, speed is  35 kmph and drawbar pull is 1533.0 kg
For current =  180 A, speed is  32 kmph and drawbar pull is 2102.0 kg